C142 B (Campbell / Callis)Chapter 2: The Components of Matter

2.1: Elements, Compounds, and Mixtures: An Atomic Overview2.2: The Observations That Led to an Atomic View of Matter2.3: The Observations That Led to the Nuclear Atom Model2.4: The Atomic Theory Today2.5: Elements: A First Look at the Periodic Table2.6: Compounds: Introduction to Bonding2.7: Compounds: Formulas, Names, and Masses2.8: Mixtures: Classification and Separation

Fig.2.1

Definitions for Components of Matter

Pure Substances - Their compositions are fixed! Elements and compounds are examples of Pure Substances.Element - Is the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means.Molecule - Is a structure that is consisting of two or more atoms that are chemically bound together and thus behaves as an independent unit.Compound - Is a substance composed of two or more elements that are chemically combined.Mixture - Is a group of two or more elements and/or compounds that are physically intermingled.

Fig.2.2

Fig2.4

Laws of Mass Conservation & Definite Composition

Law of Mass conservation: The total mass of substances does not change during a chemical reaction.

Law of Definite ( or constant ) composition: No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass.

Like Sample Problem 2.1Chemical analysis of a 9.07 g sample of calcium phosphate shows that it contains 3.52 g of Ca. How much Ca could be obtained from a 1.000 kg sample?

Mass fraction Ca =

Mass Ca in 1.000 kg =

Like Sample Problem 2.1Chemical analysis of a 9.07 g sample of calcium phosphate shows that it contains 3.52 g of Ca. How much Ca could be obtained from a 1.000 kg sample?

Mass fraction Ca = 3.52 g Ca / (9.07 g total) = 0.388(i.e., 38.8% Ca by mass in ANY sample of compound)

Mass Ca in 1.000 kg = (1.000 kg total)x(0.388 g Ca/ g total) = 0.388 kg Ca

= 388 g Ca

Fig.2.5

Mass Fraction and Mass %

Mass of Red Balls =

Mass Fraction Red = Mass % Red =

Mass Fraction Purple = Similarly, mass fraction yellow =

Check:

Mass Fraction and Mass %

Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g

Mass Fraction Red = 9.0 g / 16.0 g total = 0.56Mass % Red = 0.56 x 100% = 56% red

Mass Fraction Purple = 2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25%Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19

Check: 56% + 25% + 19% = 100%

Mass Percent Composition of Na2SO4

Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen

Elemental masses

2 x Na = 2 x 22.99 = 45.981 x S =

4 x O =

Percent of each Element

% Na = Mass Na / Total mass x 100%% Na =

% S = Mass S / Total mass x 100%% S =

% O =% O =

Check

% Na + % S + % O = 100%

Mass Percent Composition of Na2SO4

Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen

Elemental masses

2 x Na = 2 x 22.99 = 45.981 x S = 1 x 32.07 = 32.07

4 x O = 4 x 16.00 = 64.00

142.05

Percent of each Element

% Na = Mass Na / Total mass x 100%% Na = (45.98 / 142.05) x 100% =32.37%

% S = Mass S / Total mass x 100%% S = (32.07 / 142.05) x 100% = 22.58%

% O = Mass O / Total mass x 100%% O = (64.00 / 142.05) x 100% = 45.05%

Check

% Na + % S + % O = 100%

32.37% + 22.58% + 45.05% = 100.00%

Calculating the Mass of an Element in a Compound: Ammonium Nitrate

Ammonium Nitrate = NH4NO3

How much Nitrogen is in 455 kg of Ammonium Nitrate?

The Formula Mass of Cpd is:

4 x H = 4 x 1.008 = 4.032 g2 x N = 2 X 14.01 = 28.02 g3 x O = 3 x 16.00 = 48.00 g

80.052 g

Therefore mass fraction N:

28.02 g Nitrogen

80.052 g Cpd =

Mass N in sample =

Calculating the Mass of an Element in a Compound: Ammonium Nitrate

Ammonium Nitrate = NH4NO3

How much Nitrogen is in 455 kg of Ammonium Nitrate?

The Formula Mass of Cpd is:

4 x H = 4 x 1.008 = 4.032 g2 x N = 2 X 14.01 = 28.02 g3 x O = 3 x 16.00 = 48.00 g

80.052 g

Therefore mass fraction N:

28.02 g Nitrogen

80.052 g Cpd = 0.35002249 g N / g Cpd

455 kg x 1000g / kg = 455,000 g NH4NO3

455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen

28.02 kg Nitrogen

80.052 kg NH4NO4

= 159 kg Nitrogen 455 kg NH4NO3 Xor:

Fig.2.7

Law of Multiple Proportions

If elements A and B react to form two compounds,the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small wholenumbers.

Example: Nitrogen Oxides I & II

Nitrogen Oxide I : 46.68% Nitrogen and 53.32% OxygenNitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen

in 100 g of each Cpd: g O = 53.32 g & 69.55 g g N = 46.68 g & 30.45 g

g O/g N = &

Ratio of elemental mass ratios =

Law of Multiple Proportions

If elements A and B react to form two compounds,the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small wholenumbers.

Example: Nitrogen Oxides I & II

Nitrogen Oxide I : 46.68% Nitrogen and 53.32% OxygenNitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen

in 100 g of each Cpd: g O = 53.32 g & 69.55 g g N = 46.68 g & 30.45 g

g O/g N = 1.142 & 2.2842.284 2

1.142 1= = Ratio of elemental mass ratios

Dalton’s Atomic Theory

1. All matter consists of tiny particles called atoms.

2. Atoms of one element cannot be converted into atoms of another element.

3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element.

4. Compounds result from the chemical combination of a specific ratio of atoms of different elements.

Scanning Tunneling Microscope Image Reveals Individual Atoms on Silicon Surface

MoS2(0001)

Fig.2.8

Next:

Structure of Atomselectrons (-) in cloud around

nucleus = protons(+) + neutrons

Expts. which led to this picture.

Neon SignsLow pressure neon gas in evacuated tube, between electrodes.

High voltage separates +, - charges in Ne atoms:+ goes to - electrode.- goes to + electrode.

Measure current between electrodes:

Proves atoms made of +, - charges.

Fig.2.9

Deflection -> mass/charge

Fig.2.11

Millikan’s Expt.1. Measured rate of droplet’s fall without voltage: gave its mass.2. Voltage across plates influenced speed, due to charge of droplet.3. Quantitative effect of voltage w/ laws of physics -> amt. of charge on droplet.4. RESULT: Different droplets had different charges, but always a multiple of same number -> elementary charge on electron: e = 1.602x10-19 coulombs (negative).5. Mass/charge ratio x e = mass of e-

Rutherford Experiment

• Alpha (I.e., subatomic) particles bombarding the atom.

• Rationale - to study the internal structure of the atom, and to know more about the mass distribution in the atom!

• Bombarded a thin Gold foil with Alpha particles from Radium.

Fig.2.12

Ernest Rutherford (1871-1937)

• Won the Nobel Prize in Chemistry in 1908

• “It was quite the most incredible event..... It was almost as if a gunner were to fire a shell at a piece of tissue and the shell bounced right back!!!!! ”

Imagine throwing balls over a house.

Most go right over to a friend on other side, just where you tried to throw them.

A few come back to your side of house.

WHY?

Fig.2.13Moving electron cloudsurrounding nucleus.Almost all the mass in the nucleus!

__________________

Notes: mass of e- tiny relative to p+, n. p+, n have same mass (almost). e-, p+ have same charge, opposite sign.

Atomic Definitions I: Symbols, Isotopes,Numbers

XA

Z

X = Atomic symbol of the element, or element symbol

A = The Mass number; A = Z + N

Z = The Atomic Number, the Number of Protons in the Nucleus(All atoms of the same element have the same no. of protons.)

N = The Number of Neutrons in the Nucleus

Isotopes = atoms of an element with the same number of protons, but different numbers of Neutrons in the Nucleus

The Nuclear Symbol of the Atom, or Isotope

Fig.2.14

Neutral ATOMS

• 51 Cr = p+ ( ), e- ( ), n ( )

• 239 Pu = p+( ), e-( ), n ( )

• 15 N = p+( ), e-( ), n( )• 56 Fe = p+( ), e-( ), n (

)• 235 U =p+( ), e-( ), n (

)

If neutral, then # e-s = # p+s.

Numbers of each particle:

Neutral ATOMS

• 51 Cr = p+ (24), e- (24 ), n ( 27)

• 239 Pu = p+(94), e-(94), n (145)

• 15 N = p+(7), e-(7), n(8)• 56 Fe = p+(26), e-(26), n

(50)• 235 U =p+(92), e-(92),

n (143)

If neutral, then # e-s = # p+s.

Numbers of each particle:

Atomic Definitions II: AMU, Dalton, 12C Std.

Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU.

Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons.

Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard.

Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

Isotopes of Hydrogen

• 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu

• 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu

• 31H (T) 1 Proton 2 Neutrons -------- ----------

The average mass of Hydrogen is 1.008 amu

• 3H is Radioactive with a half life of 12 years.

• H2O Normal water “light water “

• mass = 18.0 g/mole , BP = 100.000000C

• D2O Heavy water

• mass = 20.0 g/mole , BP = 101.42 0C